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Description:
A matrix is Toeplitz if every diagonal from top-left to bottom-right has the same element.Now given an M x N
matrix, return True
if and only if the matrix is Toeplitz.
demo: Input: matrix = [ [1,2,3,4], [5,1,2,3], [9,5,1,2] ] Output: True Explanation: In the above grid, the diagonals are: "[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]". In each diagonal all elements are the same, so the answer is True.
描述:给定一个矩阵,判断从左上角到右下角的元素是不是相等。
思路:个人感觉关于矩阵的算法题都是找规律的,如果能够发现题目的规律,其实就很好解决了。怎么算是对角元素呢,可以发现,属于同一对角线的元素,其对角元素的(行索引-列索引)=常数,发现这个规律后就好解决了,只要判断矩阵中(行索引-列索引)属于同一个值的元素的值相同即可。
代码如下:
class Solution {
public:
bool isToeplitzMatrix(vector<vector<int>>& matrix) {
int m = matrix.size();
int n = matrix[0].size();
map<int, int> hash;
for(int i = 0; i < m;i++){
for(int j =0;j<n;j++){
if(hash.find(i-j)==hash.end()){
hash[i-j] = matrix[i][j];
}
else{
if(matrix[i][j]!=hash[i-j]){
return false;
}
}
}
}
return true;
}
};